Question

Which of the following solutions is a buffer? Check all that apply. Check all that apply. a solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M NaOH a solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M HCl a solution made by mixing 100 mL of 0.100 M KClO and 50 mL of 0.100 M KCl a solution made by mixing 100 mL of 0.100 M HClO and 500 mL of 0.100 M NaOH

Answer 1

A solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M NaOH IS A BUFFER

A solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M HCl IS NOT A BUFFER

A solution made by mixing 100 mL of 0.100 M KClO and 50 mL of 0.100 M KCl IS NOT A BUFFER

A solution made by mixing 100 mL of 0.100 M HClO and 500 mL of 0.100 M NaOH IS NOT A BUFFER

Step-by-step explanation:

A buffer is a mixture of a solution a weak acid and its conjugate base, or vice versa.

Thus, a solution by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M NaOH IS a buffer because: HClO + NaOH → NaClO. Here, weak acid is HClO and its conjugate base is NaClO

A solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M HCl IS NOT a buffer because HClO + HCl produce Cl₂ that is a gas. You don't produce the cojugate base

A solution made by mixing 100 mL of 0.100 M KClO and 50 mL of 0.100 M KCl IS NOT a buffer because these salts will not produce HClO.

a solution made by mixing 100 mL of 0.100 M HClO and 500 mL of 0.100 M NaOH IS NOT a buffer because here you have NaOH in excess. That means that all HClO will produce NaClO. In the solution you will have just the conjugate base and not its weak acid.

I hope it helps!

Answer 2

A solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M NaOH.

Step-by-step explanation:

The options answers are:

A) A solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M NaOH.

B) A solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M HCl.

C) A solution made by mixing 100 mL of 0.100 M KClO and 50 mL of 0.100 M KCl.

D) A solution made by mixing 100 mL of 0.100 M HClO and 500 mL of 0.100 M NaOH.

We have to keep in mind that a buffer solution must contain and acid and this conjutaged acid in the same solution:

`HA~<->~A^-~+~H^+`

In option B we only have acids, HClO and HCl. Is not possible to have a buffer system with these compounds

In option C we have only salts KClO and KCl. Is not possible to have a buffer system with these compounds also.

For A and B options we have an acid HClO and a base NaOH. So, the conjugate base of the acid in these cases would be produced by the addition of NaOH. So, the question is which one of these possibilities is a real buffer system.

When we check the moles presents in D we find that we have 0.01 mol of HClO and 0.05 mol NaOH, so we will have an excess of base. This excess of base will convert all the acid into his conjugated base and we need both compounds in the buffer system, the acid, and the conjugated base. So, D is not an option.